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In Gaussian plane draw solution of equation |z-(1+2i)|=2

My solution: my sol.

Wolfram solution:

wolfram sol.

I don't understand, why my solution is not right. Could anyone help me, please?

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It's supposed to be a circle. Likely your input to Wolfram Alpha was wrong. –  J. M. May 7 '12 at 11:37
    
@JM: You would think so, but you'd be wrong. –  anon May 7 '12 at 11:41
    
Your answer iscorrect. Wolfram|Alpha's is wrong. –  MJD May 7 '12 at 13:30
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2 Answers

up vote 3 down vote accepted

For some reason, W|A seems to be depicting $(1+2i)\pm2$. Now in the reals the solutions to the equation $|z-a|=b$ are $z=a\pm b$, but not in $\Bbb C$, so it might have been conflating it with that.

Your depiction of the solution set is perfectly correct.

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Tried it out; apparently Alpha does suck at plotting curves in complex format... oh well. –  J. M. May 7 '12 at 12:26
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$|z-(1+2i)|=2 \iff z-(1+2i)=2 e^{i\theta}$ for all $\theta \in\mathbb {R}$

$\iff z=1+2i+2e^{i\theta}$ for all $\theta\in\mathbb{R}$

When we take $\theta = 0, -\pi, -\frac{\pi}{2}$ and $\frac{\pi}{2}$,

we have $z=3+2i$, $-1+2i$, $1$ and $1+4i$.

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